Mary typed a 6-digit number, but the two 1’s that she typed did not show. What appeared was 2005. How many different 6-digit numbers could she have typed?
There are 4 history books and 3 math books on a shelf. How many ways can the books be arranged if the math books must be kept together?
How many possible values can there be for three coins selected from among pennies, nickels, dimes, and quarters?
Mr. and Mrs. Zimmerman want to name their baby Zimmerman so that its monogram (first, middle, and last initials) will be in alphabetical order with no letters repeated. How many such monograms are possible?
The tips of a five-pointed star are to be painted red, white, and blue. How many ways can this be done if no adjacent points can be the same color?
The 2 positions for the 1’s can be chosen from the 6 spots available in C(6, 2) = 15 ways. Since the 2,0,0,5 occupy the remaining spots in one way, the final answer is C(6, 2) = 15.
Bundling the 3 math books together, we have 4 + 1 = 5 "books" we can arrange in 5! ways. Within the bundle, the 3 math books can be arranged in 3! ways. So the total number of arrangements is 5! 3! = 120·6 = 720 ways.
There are 20 ways to choose 3 coins out of the 4 types specified, and it turns out that they all have distinct sums.
This is the number of ways to choose a subset of two letters from {A, B, … , Y}, i.e. C(25, 2) = 300.
We note that the only way to do this is ABCBC, where A, B, and C are red, white, and blue in some order. First, we choose A in C(3,1) = 3 ways, then the unique point to be colo(u)red A in C(5,1) = 5 ways. The remaining four points can now be colo(u)red using the other two colo(u)rs in two ways. So the total number of ways is 3·5·2 = 30.
July 17th, 2010 at 2:48 pm
The 2 positions for the 1’s can be chosen from the 6 spots available in C(6, 2) = 15 ways. Since the 2,0,0,5 occupy the remaining spots in one way, the final answer is C(6, 2) = 15.
Bundling the 3 math books together, we have 4 + 1 = 5 "books" we can arrange in 5! ways. Within the bundle, the 3 math books can be arranged in 3! ways. So the total number of arrangements is 5! 3! = 120·6 = 720 ways.
There are 20 ways to choose 3 coins out of the 4 types specified, and it turns out that they all have distinct sums.
This is the number of ways to choose a subset of two letters from {A, B, … , Y}, i.e. C(25, 2) = 300.
We note that the only way to do this is ABCBC, where A, B, and C are red, white, and blue in some order. First, we choose A in C(3,1) = 3 ways, then the unique point to be colo(u)red A in C(5,1) = 5 ways. The remaining four points can now be colo(u)red using the other two colo(u)rs in two ways. So the total number of ways is 3·5·2 = 30.
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